1. 1 ECE 102 Engineering Computation Chapter 20 Source Transformations Dr. Herbert G. Mayer, PSU Status 9/2/2015 For use at CCUT Fall 2015

2. 2 Syllabus Goal Goal CVS With R p Removed CVS With R p Removed CCS With R s Removed CCS With R s Removed CVS to CCS Transformation CVS to CCS Transformation Detailed Sample Detailed Sample Conclusion Conclusion Exercises Exercises

3. 3 Goal The Node-Voltage and Mesh-Current Methods are powerful tools to compute circuit parameters Cramer’s Rule is especially useful for a large number of unknowns Sometimes a circuit can be transformed into another one that is simpler, yet electrically equivalent Generally that will simplify computations We’ll learn a few source transformations here Method 1: remove parallel load from CVS Method 2: remove serial load from CCS Method 3: Transform CVS CCS bilaterally

4. 4 CVS With R p Removed

5. 5 Removing the load R p parallel to the CVS has no impact on externally connected loads R L Such loads R L —not drawn here— will be in series with resistor R Removal of R p decreases the amount of current that the CVS has to produce, to deliver equal voltage to both R p and the series of R and load R L This simplification is one of several source transformations an engineer should look for, before computing unknowns in a circuit

6. 6 CCS With R s Removed

7. 7 Removing the load R s in series with the CCS has no impact on external loads R L Such a load R L —not drawn here— will be parallel to resistor R Removal of R s will certainly decrease the amount of voltage the CCS has to produce, to deliver equal current to both R s in series with R parallel to R L Such a removal is one of several source transformations to simplify computing unknown units in a circuit

8. 8 CVS to CCS Bilateral Transformation

9. 9 CVS to CCS Transformation A given CVS of V s Volt with resistor R in series produces a current i L in a load, connected externally That current also flows through connected load R L i L = V s / ( R + R L ) A CCS of i S Ampere with parallel resistor R produces a current i L in an externally connected load R L For the transformation to be correct, these currents must be equal for all loads R L i L = i s * R / ( R + R L ) Setting the two equations for i L equal, we get: i s = V s / R V s = i s * R

10. 10 Detailed Sample We’ll use these simplifications in the next example to generate an equivalent circuit that is minimal I.e. eliminate all redundancies from right to left This example is taken from [1], page 110-111, expanded for added detail First we analyze the sample, identifying all # of Essential nodes ____ # of Essential branches ____ Then we compute the power consumed or produced in the 6V CVS

11. 11 Detailed Sample, Step a

12. 12 Detailed Sample identify all: # of Essential nodes __4__ # of Essential branches __6__

13. 13, Detailed Sample, Step b

14. 14, Detailed Sample, Step c

15. 15, Detailed Sample, Step d

16. 16, Detailed Sample, Step e

17. 17, Detailed Sample, Step f

18. 18, Detailed Sample, Step g

19. 19, Detailed Sample, Step h

20. 20 Power in 6 V CVS The current through network h, in the direction of the 6 V CVS source is: i = ( 19.2 - 6 ) / ( 4 + 12 ) [ V / Ω ] i = 0.825 [ A ] Power in the 6 V CVS, being current * voltage is: P = P 6V = i * V = 0.825 * 6 P 6V = 4.95 W That power is absorbed in the 6 V source, it is not being delivered by the 6 V source! It is delivered by the higher voltage CVS of 19.2 V

21. 21 Conclusion Such source transformations are not always possible Exploiting them requires that there be a certain degree of redundancy Frequently that is the case, and then we can simplify Engineers must check carefully, how much simplification is feasible, and then simplify But no more

22. 22 Exercise 1 Taken from [1], page 112, Example 4.9, part a) Given the circuit on the following page, compute the voltage drop v 0 across the 100 Ω resistor Solely using source transformations Do not even resort to KCL or KVL, just simplify and then use Ohm’s Law

23. 23 Exercise 1

24. 24 Exercise 1 We know that the circuit does not change, when we remove a resistor parallel to a CVS Only the power delivered by the CVS will change So we can remove the 125 Ω resistor We also know that the circuit does not change, when we remove a resistor in series with a CCS Only the overall power delivered by the CCS will change So we can remove the 10 Ω resistor

25. 25 Exercise 1, Simplified Step 1

26. 26 Exercise 1, Cont’d Computation of v 0 does not change with these 2 simplifications If we substitute the 250 V CVS with an equivalent CCS, we have 2 parallel CCS These 2 CCSs can be combined

27. 27 Exercise 1, Simplified Step 3

28. 28 Exercise 1, Cont’d Combine 2 parallel CCS of 10 A and -8 A And combine 3 parallel resistors: 25 || 100 || 20 Ω = 10 Ω Yielding an equivalent circuit that is simpler, and shows the desired voltage drop v 0 along the equivalent source, and equivalent resistor

29. 29 Exercise 1, Simplified Step 2

30. 30 Exercise 1, Cont’d We can compute v 0 v 0 = 2 A * 10 Ω v 0 = 20 V

31. 31 Exercise 2, Compute Power of V 250 Next compute the power p s delivered (or if sign reversed: absorbed) by the 250 V CVS The current delivered by the CVS is named i s And it equals the sum of i 125 and i 25

32. 32 Exercise 2, Compute Power of V 250

33. 33 Exercise 2, Compute Power of V 250 i s = i 125 + i 25 i s = 250/125 + (250 - v 0 )/25 i s = 250/125 + (250 - 20)/25 i s = 11.2 A Power p s is i * v p s = 250 * 11.2 = 2,800 W

34. 34 Exercise 3, Compute Power of 8 A CCS Next compute the power p 8A delivered by the 8 A CCS First we find the voltage drop across the 8 A CCS, from the top essential node toward the 10 Ω resistor, named v 8A The voltage drop across the 10 Ω resistor is simply 10 Ω * the current, by definition 8 A, named v 10Ω That is v 10Ω = 80 V

35. 35 Exercise 3, Compute Power of 8 A CCS

36. 36 Exercise 3, Compute Power of 8 A CCS V 0 = v 8A + v 10Ω V 0 = 20 V 20= 8*10 + v 8A v 8A = 20 - 80 v 8A = -60 Power p 8A is i 8A * v 8A p 8A = i 8A *v 8A